Question 172039
A game is played using one die. If the die is rolled and shows a 2, the player wins $8. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value?
<pre><font size= 4 color= "indigo"><b>

There are two possibilities.  

1. He wins $7, ($8 - $1 to play), when a 2 is thrown 
with probability {{{1/6}}}.

2. He loses $1, considered as a "negative win", or -$1, 
when a 1,3,4,5,or 6 is thrown, with a probability of {{{5/6}}}.


{{{matrix(2,6,

  x,  "|", 7,   "|",  -1,  "|",
P(x), "|", 1/6, "|", 5/6,  "|")}}}

{{{E(x) = sum((x*P(x)))=(7)(1/6)+(-1)(5/6)=7/6-5/6=2/6=1/3}}}

So, the expected value is 

{{{1/3}}}rd of {{{matrix(1,1,"$1")}}}, or {{{matrix(1,2,33&1/3,cents)}}}

That means if you played the game over and over many times,
you would expect to average winning {{{matrix(1,2,33&1/3,cents)}}} per
game.

Edwin</pre>